DESIGN OF HYDRAULIC STRUCTURES

ALLUVIAL CHANNEL DESIGN

KENNEDY’S THEORY

R. G. Kennedy – 1895

Non-silting non-scouring reaches for 30 years in Upper Bari Doab Canal (UBDC) system.

Vertical eddies generated from the bed are responsible for keeping silt in suspension.

Critical velocity

Mean velocity which keeps the channel free from silting and scouring.

Vo = 0.55 D0.64

which can be written in general form as,

Vo = C Dn

where, Vo = critical velocity, D = depth of water

C = constant and n = index number

Later on realizing the channel material (sandy silt in UBDC), he modified the equation as

Vo = 0.55 m D0.64

or Vo = C m Dn

where,

m = C.V.R = Critical Velocity Ratio = V/Vo ; V = actual velocity

m = 1.1 – 1.2 coarser sand

m = 0.7 – 0.9 finer sand

m = 0.85 Sindh canals

Values of C (the constant in Kennedy’s Eq.) and m (the Critical Velocity Ratio, CVR) for various grades of silt

Type of silt grade C m

Coarser silt 0.7 1.3

Sandy loam silt 0.65 1.2

Coarse light sandy silt 0.59 1.1

Light sandy silt 0.53 1.0

Rugosity coefficient

Kennedy used Kutters equation for determining the mean velocity of flow in the channel

Where N depends upon the boundary material

Channel condition N

Very good 0.0225

Good 0.025

Indifferent 0.0275

Poor 0.03

Discharge (cumec) N (in ordinary soil)

14 – 140 0.025

140 – 280 0.0225

> 280 0.02

RS

R

N

S

SNV

00155.0

231

00155.0123

Water Surface Slope

No relationship by Kennedy.

Governed by available ground slope.

Different sections for different slopes.

Wood’s normal design table for B/D ratio.

Silt Carrying Capacity of Channel

Qt = K B Vo0.25

where

Qt = total quantity of silt transported

B = bed width

Vo = critical velocity

K = constant, whose value was not determined by Kennedy

Modification for Sindh Canals

In 1940, while designing Guddu Barrage project canals, K. K. Framji

proposed B/D ratio for Sindh canals as:

5.15.3 61 QD

B

Discharge

m3/sec

B/D ratio for

standard section

B/D ratio for

limiting section

100 6.0 4.0

1000 8.4 5.0

5000 13.0 8.0

Design Procedure

Case I : Given Q, N, m and S (from L-section)

1. Assume D

2. Calculate velocity from Kennedy’s equation, VK = 0.55 m D0.64

3. Calculate area, A = Q / VK

4. Calculate B from A = B D + z D2 ; assume side slope 1(V) : ½(H), if not given.

5. Calculate wetted perimeter and hydraulic mean depth from;

6. Determine mean velocity from Chezy’s equation, Vc =C √(RS)

if Vc = Vk then O.K.

otherwise repeat the above procedure with another value of D until Vc = Vk.

Note: increse D if Vk < Vc

decrease D if Vk > Vc

DBP 5

DB

DBD

P

AR

5

5.0 2

Problem:

Design an irrigation channel for the following data using Kennedy’s

theory:

Full Supply Discharge (F.S.Q) = 14.16 cumec

Slope, S = 1/5000

Kutter’s rugosity coefficient, N = 0.0225

Critical velocity ratio, m =1

Side slope, z = ½

Solution:

1. Assume D = 1.72 m

2. Vk = 0.55 m D0.64 =0.55(1)(1.72)0.64 = 0.778 m

3. A = Q/Vk = 14.16/0.778 = 18.2 m2

4. A = B D + 0.5 D2 for z =1/2 or 0.5

18.2 = 1.72 B + 0.5(1.72)2

B = 9.72 m

5.R = A / P = 18.2 / 13.566 = 1.342 m

6.

Vc = 0.771 m

≈ 0.778 m

Result:B = 9.72 m

D = 1.72 m

m 566.13)72.1(572.95 DBP

RS

R

N

S

SNVc

00155.0

231

00155.0123

50001342.1

342.1

0225.0

50001

00155.0231

50001

00155.0

0225.0

123

cV

EXAMPLE PROBLEM

Q = 80 m3/sec

S = 1:5500 = 0.00018 m/m

m = 1

DATA

Assume D = 2.5 m

Vk = 0.55 D0.64 = 0.989 m/sec

A = 80.918 m2

Side Slope = 1V:1.5H

n = 0.0225

A = B D+ 1.5D2

B = 28.617 m

P = 32.223 m

R = A/ P = 2.511 m

Using Kutter’s Formula in S.I. Units

C = 52.479

Vc = C√RS = 1.121 m/sec

Keep on trailing till Vc = Vk

D

B

1

1.5

1.803

PROBLEM

Design an irrigation channel using Kennedy’s theory that irrigates

a cultral commanded area of 50,000 hectares. Given that delta of

the crops is 0.12 m/month, slope of country is 1 in 4000 and

Kutter’s rugosity coefficient is 0.023. Assume the missing data.

222 5.05.0 DyDDBDA

)1()5.0(2 yDA

)55.0).(5.0( 64.02 mDyDQ

64.21

5.055.0

ym

QD

DB

DBDR

5

5.0 2

Problem:

Using Kennedy’s theory design an irrigation channel to carry a discharge of 56.63 cumec. Assume N = 0.0225, m = 1.03 and B/D = 11.3.

Solution:

1. B/D = 11.3, therefore B = 11.3 D

A = B D + 0.5 D2 =11.3 D2 + 0.5 D2 = 11.8 D2

2. V = 0.55 m D0.64 = 0.55 (1.03) D0.64 = 0.5665 D0.64

3. Q = A V

56.63 = (11.8 D2 ) (0.5665 D0.64 )

D = 2.25 m

4. B = 11.3 (2.25) = 25.43 m

5. R = A / P

A = B D + 0.5 D2 = (25.43)(2.25) + 0.5 (2.25)2 = 59.75 m2

P = B + √5 D = 25.43 + √5 (2.25) = 30.46 m

R = 59.75 / 30.46 = 1.96 m

6. V = 0.55 m D0.64 = 0.55 (1.03) (2.25)0.64 = 0.95 m/sec

7.

Simplifying, we get;

67.44 S3/2 – 0.93 S + 1.55x10-3 S1/2 = 1.68x105

Solving by trial and error, we get

S = 1 in 5720

Results:B = 25.43 m

D = 2.25 m

S = 1 / 5720

RS

R

N

S

SNV

00155.0

231

00155.0123

S

S

S )96.1(

96.1

0225.000155.0231

00155.0

0225.0

123

95.0

Case III : Given S, N, m and B/D

1. From the B/D ratio, determine B in terms of D.

2. Determine A, P and R in terms of D.

3. From Kennedy’s equation, determine velocity (Vk) in terms of D.

4. Putting values of N, S and R in the Chezy’s equation and Kutter’s

formula, determine velocity (Vc). Simplify the expression, and solve it

by trail and error for D.

5. Knowing D, calculate B, A and Vk.

6. Using continuity equation, determine the discharge (Q).

Problem:

Design a section by Kennedy’s theory, given B/D = 5.7, S = 1/(5000+X)

and N = 0.0225. Also determine the discharge carried by the channel.

Solution:

B/D = 5.7, B = 5.7 D

Assuming z = ½

Since V = 0.55 m D0.64

Assuming m =1

V = 0.55 D0.64 ---------- (1)

Also

DD

D

DD

DD

DB

DBDR 78.0

94.7

2.6

57.5

5.07.5

5

5.0 2222

RS

R

N

S

SNV

00155.0

231

00155.0123

2783.0

939.05000178.0

78.0

0225.0

50001

00155.0231

50001

00155.0

0225.0

123

D

DD

D

V

Equating equation (1) and (2)

0.55 D1.14 – 0.939 D + 0.43 D0.64 = 0

By trial and error

D = 2.1 m

B = 5.7 x 2.1 = 11.97 m

A = B D + z D2 = (11.97 x 2.1) + 0.5 (2.1)2 = 14.175 m2

V = 0.55 (2.1)0.64 = 0.884 m/sec

Q = A V = (14,175)(0.884) = 12.53 m3/sec.

Results:

B = 11.97 m

D = 2.1 m

Q = 12.53 cumec

D

DD

783.0

939.055.0 64.0

Shortcomings of Kennedy’s theory

1. The method involves trial and error.

2. Shape of section i.e. B/D is not known inadvance.

3. Kutter’s equation is used instead of Manning’sequation. Therefore limitations of Kutter’sformula are also incorporated in Kennedy’stheory. Moreover it involves morecomputations.